GATE PH 2013 • Q49 MCQ • 2 Marks

Section 5: Thermodynamics and Statistical Physics

Statistical Ensembles & Partition Functions

Subtopic: Partition Function & Thermodynamic Connection

Question Statement:

There are four energy levels E, 2E, 3E and 4E (where E > 0). The canonical partition function of two distinguishable particles is (A) e^(-5βE) (e^(2βE) + e^(βE) + 2 + e^(-βE) + e^(-2βE)) (B) (e^(-βE) + e^(-2βE) + e^(-3βE) + e^(-4βE))^2 (C) 1/2 (e^(-βE) + e^(-2βE) + e^(-3βE) + e^(-4βE))^2 (D) e^(-8βE) (1 + e^(-βE) + e^(-2βE))^2

Options:

  • (A) e^(-5βE) (e^(2βE) + e^(βE) + 2 + e^(-βE) + e^(-2βE))
  • (B) (e^(-βE) + e^(-2βE) + e^(-3βE) + e^(-4βE))^2
  • (C) 1/2 (e^(-βE) + e^(-2βE) + e^(-3βE) + e^(-4βE))^2
  • (D) e^(-8βE) (1 + e^(-βE) + e^(-2βE))^2

Official Answer Key:

B

Step-by-Step Solution & Explanation

### **Official Solution & Step-by-Step Derivation (GATE PH 2013 Q.49)** **1. Statistical Physics of Distinguishable Particles**: - For non-interacting distinguishable particles, each particle independently accesses any of the available single-particle states without restriction. - Single-particle partition function $z_1$: $z_1 = \sum_{i=1}^4 e^{-\beta \epsilon_i} = e^{-\beta E} + e^{-2\beta E} + e^{-3\beta E} + e^{-4\beta E}$ **2. Two-Particle Partition Function**: - For $N = 2$ distinguishable particles: $Z = (z_1)^2 = \left( e^{-\beta E} + e^{-2\beta E} + e^{-3\beta E} + e^{-4\beta E} \right)^2$ - This corresponds precisely to **Option (B)**. **3. Official Verification**: The verified official IIT Bombay Master Answer is **Option (B)**.

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